The proof of theorem 5 of Some further results on ideal convergence in topological spaces by Pratulananda Das is unclear to me.
Theorem 5. For any sequence $x = (x_n)_{n\in\mathbb N}$ in a hereditarily Lindelöf space $X$, there exists a sequence $y = (y_n)_{n\in\mathbb N}$ in $X$ such that $\{n \in\Bbb N: x_n \neq y_n\} \in \mathcal{I}$ and $\mathcal{I}(C_x) = L(y)$ provided $\mathcal{I}$ satisfies the condition $(AP)$ or in other words is a $P$-ideal.
Proof. The proof is finished if $\mathcal{I}(C_x) = L(x)$.
- How so? It only gives that $I(C_y)=L(y)$. How do we know $L(x)=L(y)$?
If not then $\mathcal{I}(C_x)\nsubseteq L(x)$.
- $\mathcal{I}(C_x)\neq L(x)$ if either or both of $\mathcal{I}(C_x)\subset L(x)$ and $L(x)\subset \mathcal{I}(C_x)$ does not hold. So maybe $\mathcal{I}(C_x)\nsubseteq L(x)$ or $L(x)\nsubseteq \mathcal{I}(C_x)$. How does he know it's the first one?
For each $z ∈ L(x)\setminus \mathcal{I}(C_x)$ we can find an open set $U_z$ containing $z$ such that $\{n \in \mathbb N: x_n \in U_z\} \in \mathcal{I}$.
- Even if $\mathcal{I}(C_x)\nsubseteq L(x)$, how can we say $z\in L(x)\setminus\mathcal{I}(C_x)$? Because $\mathcal{I}(C_x)\nsubseteq L(x)\implies \mathcal{I}(C_x)\setminus L(x)\neq \varnothing$ and not $L(x)\setminus\mathcal{I}(C_x)\neq \varnothing$ because $\mathcal{I}(C_x)\nsubseteq L(x)$ holds when $\mathcal{I}(C_x)$ is a proper superset of $L(x)$ and in that case $L(x)\setminus\mathcal{I}(C_x)=\varnothing$ and we shall find no $z\in L(x)\setminus\mathcal{I}(C_x)$
Now $\{U_z: z \in L(x)\setminus \mathcal{I}(C_x)\}$ forms an open cover of $L(x)\setminus\mathcal{I}(C_x)$ and so it has a countable subcover $\{U_{z_i}: i \in \mathbb N\}$ (say). Put $A_i = \{n \in \mathbb N: x_n \in U_{z_i}\}$. Then $(A_i)$ is a sequence of sets in $\mathcal{I}$ and since $\mathcal{I}$ has property (AP), there exists an $A \in\mathcal I$ such that $A_i\setminus A$ is finite for each $i$. If $\mathbb N\setminus A = \{k_n: n \in \mathbb N\}$, then construct $y = (y_n)_{n\in\mathbb N}$ as follows:
$$\begin{cases} y_n=x_{k_n}&\text{if }n\in A,\\ y_n=x_n&\text{if }n\notin A. \end{cases}$$
Clearly then $\{n \in \mathbb N: x_n \neq y_n\} \subset A$ and so belongs to $\mathcal I$. By Theorem 4 we have $\mathcal{I}(C_x) = \mathcal{I}(C_y)$. But note that the subsequence $(x_{k_n} )$ of $(y_n)_{n\in\mathbb N}$ has no accumulation point in $L(x)\setminus \mathcal{I}(C_x)$ and so has no $\mathcal{I}$-limit point of $(y_n)$ (since $\mathcal I$ is admissible).
- I cannot see at all how it follows that if the subsequence $(x_{k_n} )$ of $(y_n)_{n\in\mathbb N}$ has no accumulation point in $L(x)\setminus \mathcal{I}(C_x)$, then it has no $\mathcal{I}$-limit point of $(y_n)$.
So $L(y) = \mathcal{I}(C_y)$ and consequently we have $L(y) = \mathcal{I}(C_x)$. $\square$
Necessary definitions and statements
In the following $X$ is a topological space, $\mathcal{I}$ is an ideal on $\mathbb N$, $x = (x_n)_{n \in \mathbb N}$ is a sequence in $X$.
$y \in X$ is called an $\mathcal{I}$-limit point of $x$ if there is a strictly increasing sequence $( m_k )_{k \in \mathbb N}$ in $\mathbb N$ such that $\{ m_k : k \in \mathbb N \} \notin \mathcal{I}$ and $\lim_{k \to \infty} x_{m_k} = y$.
$y \in X$ is called an $\mathcal{I}$-cluster point of $x$ if for every open set $U$ containing $y$, $\{n \in N: x_n \in U\}\notin\mathcal{I}$. And the collection of all $\mathcal{I}$-cluster points of $X$ is denoted by $\mathcal{I}(C_x)$.
$z\in L(x)$ if for each neighborhood $W$ of $z$, $\{n \in \mathbb N: x_n \in W \}$ is infinite (i.e., $L(x)$ is the set of all cluster points of $x$).
$\mathcal{I}$ has property (AP) (or is a P-ideal) if for every sequence $( A_n )_{n \in \mathbb N}$ in $\mathcal I$ there exists an $A_\infty \in \mathcal I$ such that $A_n \setminus A_\infty$ is a finite set for every $n \in \mathbb N$.
$\mathcal{I}$ is admissible (or free) if it contains all singletons.
Theorem 4. If $x = (x_n)_{n \in \mathbb N}$ and $y = (y_n)_{n \in \mathbb N}$ are two sequences in $X$ such that $\{ n \in \mathbb N : x_n \neq y_n \} \in \mathcal I$, then $\mathcal{I} ( C_x) = \mathcal{I} ( C_y )$.
(The paper actually states "... $\{ n \in \mathbb N : x_n \neq y_n \} \notin \mathcal I$ ...", however following the proof this is clearly a typo.)
Everything I am doing in this answer is under the assumption that the ideal $\mathcal I$ is admissible.
We want to show that there exists $y$ such that $\mathcal I(C_x)=L(y)$. If $I(C_x)=L(x)$, then we can simply choose $y=x$.
This is clarified here: What is the usual relationship between $L(x)$ and $I(C_x)\ ?$
We can show that for an admissible ideal we have $\mathcal I(C_x)\subseteq L(x)$. So if the two sets are different, then $L(x)\nsubseteq \mathcal I(C_x)$. (The proof in the paper writes it in the opposite way, which is a typo.) And we have in such case that $L(x)\setminus \mathcal I(C_x)\ne\emptyset$.
In this part of the proof the author is going to show that something holds for every such $z$. So, naturally, the proof starts with assuming that $z\in L(x)\setminus \mathcal I(C_x)$. Then they get existence of $U_z$ with the properties described in the proof for each $z\in L(x)\setminus\mathcal{I}(C_x)$.
Let us try to repeat the last part of the proof it detail so that we clearly see all necessary components of the proof.
We have an open over $\{U_{z_i}; i\in\mathbb N\}$ of $L(x)\setminus \mathcal I(C_x)$. For each $i$ we denote $A_i=\{n\in\mathbb N; x_n\in U_{z_i}\}$. Using the fact that $\mathcal I$ is P-ideal we get a set $A\in\mathcal I$ such that each $A_i\setminus A$ is finite.
Then we denote $\mathbb N \setminus A=\{k_1<k_2<\dots<k_n<\dots\}$ and we define $$ y_n= \begin{cases} x_{k_n} & \text{if }n\in A, \\ x_n & \text{of }n\notin A. \end{cases} $$ Since the sequence $x$ and $y$ differ only on the set $A$ which belongs to the ideal $\mathcal I$, we get that $\mathcal I(C_x)=\mathcal I(C_y)$.
We want to prove that $L(y)=\mathcal I(C_x)$. We will do this in several steps.
If some subsequence of $y$ converges to $z$, then we can get either subsequence of $x_n$ or subsequence of $x_{k_n}$ which converges to $z$. In either case $z\in L(x)$.
If $x_{k_n}\in U_{z_i}$, then $k_n\in A_i \cap (\mathbb N\setminus A)= A_i\setminus A$. So there is only finitely many such $k_n$'s. And since $n\mapsto k_n$ is bijective, we also have only finitely many $n$'s.
We have $$\{n\in\mathbb N; x_n\in U_{z_i}\} \subseteq \{n\in\mathbb N; x_{k_n}\in U_{z_i}\} \cup (\{n\in\mathbb N; n\in U_{z_i}\}\cap(\mathbb N\setminus A)).$$ The first set is finite by Observation 2. The second one is precisely $A_i\setminus A$, which is finite too.
(Notice that this really implies that: "$(x_{k_n})$ of $(y_n)_{n\in\mathbb N}$ has no accumulation point in $L(x)\setminus \mathcal{I}(C_x)$" (as claimed in the paper). However, for me the proof seemed to be a bit clearer when I formulated it as above.)
If $z\in L(x)\setminus\mathcal I(C_x)$, then there is some $i\in\mathbb N$ such that $z\in U_{z_i}$. (Here we are using simply the fact that $\{U_{z_i}; i\in\mathbb N\}$ is covering of $L(x)\setminus\mathcal I(C_x)$.)
Then by observation 3 we have only finitely many terms of the sequence $y$ in this neighborhood of $z$ and therefore $z$ is not a cluster point of $y$.
Conclusion. From $L(y)\subseteq L(x)$ and $L(y)\cap (L(x)\setminus\mathcal I(C_x))=\emptyset$ we get that $$L(y) \subseteq \mathcal I(C_x).$$ We also know that $$\mathcal I(C_y)\subseteq L(y)$$ since $\mathcal I$ is admissible ideal. (See here.)
Using this, together with the fact that $\mathcal I(C_y)=\mathcal I(C_x)$, we get $$\mathcal I(C_y) \subseteq L(y)\subseteq \mathcal I(C_x)=\mathcal I(C_y)$$ which means that all these sets are equal to each other. In particular, we get $$L(y)=\mathcal I(C_x).$$
It is probably worth mentioning that this results is a generalization of Theorem 2 from the paper J. A. Fridy: Statistical limit points, Proc. Amer. Math. Soc. 118 (1993), 1187-1192; https://doi.org/10.1090/S0002-9939-1993-1181163-6 http://www.jstor.org/stable/2160076 In this paper the same result is shown for the ideal of sets having asymptotic density zero. The proof of the more general result seems to be along the same lints as Fridy's proof of the special case.